JEE Physics 2025 Question with Solution

Given: $i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right)$

Find: The RMS value of the current.

The current has two parts:

• a DC component, $I_{\text{DC}} = 5\sqrt{2}\, \text{Amp}$
• an AC component, $i_{\text{AC}}(t) = 10\cos\left(650\pi t + \frac{\pi}{6}\right)\, \text{Amp}$

For a sinusoidal current with peak value $I_{\text{peak}}$, the RMS value is

$$I_{\text{AC,rms}} = \frac{I_{\text{peak}}}{\sqrt{2}}$$

So,

$$I_{\text{AC,rms}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}\, \text{Amp}$$

When DC and AC components are both present, the total RMS value is

$$I_{\text{rms}} = \sqrt{I_{\text{DC}}^2 + I_{\text{AC,rms}}^2}$$

Substituting the values,

$$I_{\text{rms}} = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2}$$ $$I_{\text{rms}} = \sqrt{50 + 50}$$ $$I_{\text{rms}} = \sqrt{100}$$ $$I_{\text{rms}} = 10\, \text{Amp}$$

Verification:

• $(5\sqrt{2})^2 = 50$
• The sum of squares is $100$
• Therefore, $\sqrt{100} = 10$

Therefore, the correct option is C and the RMS value is $10\, \text{Amp}$.

Given: $i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right)$

Find: The RMS value of the current.

Start from the definition through squaring the current:

$$i^2 = \left(5\sqrt{2} + 10\cos\left(650\pi t + \frac{\pi}{6}\right)\right)^2$$

Expanding,

$$i^2 = 50 + 100\cos^2\left(650\pi t + \frac{\pi}{6}\right) + (2)(5\sqrt{2})(10)\cos\left(650\pi t + \frac{\pi}{6}\right)$$

Over a complete cycle, the average value of $\cos\left(650\pi t + \frac{\pi}{6}\right)$ is $0$ and the average value of $\cos^2\left(650\pi t + \frac{\pi}{6}\right)$ is $\frac{1}{2}$. Hence,

$$\langle i^2 \rangle = 50 + 100\left(\frac{1}{2}\right) + 0$$ $$\langle i^2 \rangle = 100$$

Therefore,

$$I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \sqrt{100} = 10\, \text{Amp}$$

Therefore, the correct option is C.

Note: the provided alternate solution states an inequality in one intermediate line, but the cycle-average method gives the exact RMS value $10\, \text{Amp}$.